How do you solve this equation using a least common denominator?

That's right. So, multiply the first fraction by (x+2)/(x+2), and you get:
3x(x+2)/(x-2)(x+2)
And the second one, you multiply by (x-2)/(x-2) to get:
1(x-2)/(x-2)(x+2)
Now, distribute and combine those:
(3x^2+6x+x-2)/(x+2)(x-2)=(3x^2+7x-2)/(x^2-4)
Now you can say 3x^2+7x-2=-4
3x^2+7x+2=0
Factor that and solve for x, but remember that x cannot equal +/-2.

Brackets should be used in question. Assume meant to read as:-
3x / (x - 2) + 1 / (x + 2) = (- 4) / (x - 2)(x + 2)

3x (x + 2) + 1 (x - 2) = (- 4)
3x² + 6x + x - 2 = - 4
3x² + 7x + 2 = 0
( 3x + 1 ) ( x + 2 ) = 0
x = - 1/3 , x = - 2

(3x/[x - 2]) + (1/[x + 2]) = (- 4/[x² - 4])
([3x² + 6x] + [x - 2])/(x² - 4) = (- 4/[x² - 4])
3x² + 7x - 2 = - 4
3x² + 7x + 2 = 0
(x + 2)(3x + 1) = 0

x + 2 = 0, x = - 2
3x + 1 = 0, 3x = - 1, x = -1/3

Answer: x = - 2, x = - 1/3

Proof (where x = - 2):
(3[- 2]/[- 2 - 2]) + (1/[- 2 + 2]) = - 4/(- 2² - 4)
(- 6/0) + (1/0) = - 4/(4 - 4)
0 + 0 = - 4/0
0 = 0

Proof (where x = - 1/3)
(3[- 1/3]/[- 1/3 - 2]) + (1/[- 1/3 + 2]) = - 4/([- 1/3]² - 4)
(- 3/[- 2 1/3]) + (1/[1 2/3]) = - 4/(1/9 - 4)
(- 3/[- 7/3]) + (1/[5/4]) = - 4/(- 3 8/9)
(- 3 * - 3/7) + (1 * 4/5) = - 4/(- 35/9)
9/7 + 4/5 = - 4 * - 9/35
36/35 = 36/35

5x/4 - 3/x = 1/4 Multiply 4 times x. That would be the least common denominator. Good luck!

yeah that's good. now, multiply the first fraction (top and bottom) by (x+2) and the second by (x-2).

then u should get 3x(x+2) / (x-2)(x+2) + 1(x-2)/(x+2)(x-2) (NOTE: this is only for the left side, don't touch the right side yet)

for the left side, after u distribute and combine like terms u should have:

(3x^2 + 7x -2) / (x+2) (x-2)

now for the right:

u can factor out the denominator using difference of squares, thus x^2 -4 = (x+2)(x-2)

for the right side u will have:

-4 / (x+2)(x-2)

now write the whole expression down:

(3x^2 + 7x -2) / (x+2) (x-2) = -4 / (x+2)(x-2)

notice that the denominators are the same, thus u can get rid of all them. now u will have:


3x^2 + 7x -2 = -4

now add 4 to both sides of the equation:

3x^2 + 7x +2 = 0

the last step is to factor this quadratic equation, thus

(3x+1)(x+2)=0

set each linear factor equal to zero, and solve

therefore, x= - 1/3 and x= -2

3x/(x - 2) + 1/(x + 2) = -4/(x^2 - 4)
3x(x + 2)/(x - 2)(x + 2) + 1(x - 2)/(x + 2)(x - 2) = -4/(x^2 - 4)
(3x^2 + 6x)/(x^2 - 4) + (x - 2)/(x^2 - 4) = -4/(x^2 - 4)
3x^2 + 6x + x - 2 = -4
3x^2 + 7x - 2 + 4 = 0
3x^2 + 7x + 2 = 0
(3x + 1)(x + 2) = 0

3x + 1 = 0
3x = -1
x = -1/3

x + 2 = 0
x = -2

∴ x = -1/3 , -2

3x(x+2) + (x-2) = -4 [deno are same so they are = 1]
3x^2+6x+x+2=0
(3x+1)(x+2)=0

x = -1/3, -2

you r right with least common denominator as (x-2)(x+2)

3x( x+ 2) + (x-2) = -4

3x^2 + 7x +2=0
factorize them , or use quadratic formula
3x^2 + 6x +x +2 = 0

3x( x+2) + 1 (x+2) = 0

3x+1 = 0 or x+2 = 0

x= -1/3 , -2

ok so you have this equation
3x/(x-2) + 1/(x+2) = -4/(x^2-4)

1. what i would do first would be move all terms to one side of the equation so you get a zero

3x/(x-2) + 1/(x+2) + 4/(x^2-4) = 0 (don't forget to change the sign)

2. factor all the denominator; you can see the third denominator (x^2 - 4) is the difference of two squares and = (x+2)(x-2)

3x/(x-2) + 1/(x+2) + 4/(x-2)(x+2) = 0

3. look at all the denominators we have now are; (x-2), (x+2), and (x+2)(x-2); the lowest common denominator, in which all these denominators can divide evenly is (x+2)(x-2)

4. multiply this LCD by every term in the equation
(x+2)(x-2)3x/(x-2) + (x+2)(x-2)1/(x+2) + (x-2)(x+2)4/(x-2)(x+2) = 0

5. simplify to get rid of denominators (the aim of the lcd); (x+2)3x + (x-2)1 + 4 = 0

6. 3x^2 + 6x + x - 2 + 4 = 0
3x^2 + 7x + 2 = 0

use the quadratic formula -b +- sqrt(b^2 - 4(ac) ) / 2a
where a = 3
b = 7
c = 2

-7 +- sqrt(49 - 4(6)) / 6
-7 +- 5 ) / 6
-2 and -1/3

maybe some of your confusion is relating to the lowest common denominator.

it is used differently in an expression than how it is used in an equation; in an expression it is used to give each term a denominator, then add or function accordingly; in an EQUATION however as you can see it is used to ELIMINATE the denominators so the numerators alone can be used. any questions [email protected]

((3x^2+6x+x-2)/x^2-4)=-4/x^2-4

NOT SURE, pls make your question more understandable